Answer: The enthalpy of formation of 
is -396 kJ/mol
Explanation:
Calculating the enthalpy of formation of
The chemical equation for the combustion of propane follows:
The equation for the enthalpy change of the above reaction is:
![\Delta H^o_{rxn}=[(2\times \Delta H^o_f_{(SO_3(g))})]-[(2\times \Delta H^o_f_{(SO_2(g))})+(1\times \Delta H^o_f_{(O_2(g))})]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B%282%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28SO_3%28g%29%29%7D%29%5D-%5B%282%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28SO_2%28g%29%29%7D%29%2B%281%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28O_2%28g%29%29%7D%29%5D)
We are given:
Putting values in above equation, we get:
![-198=[(2\times \Delta H^o_f_{(SO_3(g))})]-[(2\times \Delta -297)+(1\times (0))]\\\\\Delta H^o_f_{(SO_3(g))}=-396kJ/mol](https://tex.z-dn.net/?f=-198%3D%5B%282%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28SO_3%28g%29%29%7D%29%5D-%5B%282%5Ctimes%20%5CDelta%20-297%29%2B%281%5Ctimes%20%280%29%29%5D%5C%5C%5C%5C%5CDelta%20H%5Eo_f_%7B%28SO_3%28g%29%29%7D%3D-396kJ%2Fmol)
The enthalpy of formation of is -396 kJ/mol
Answer:
The answer to your question is B. metallic
Explanation:
Covalent bonding is a bond between two nonmetals and the difference in electronegativity is between 0 and 1.7. Sodium could not have this kind of bond because is a metal.
Ionic bonding is a bond between a metal and a nonmetal and the difference of electronegativity is higher than 1.7. Sodium can have this kind of bond it is necessary one nonmetal.
Metallic, sodium has a metallic bond because this bond is characteristic of metals.
Answer:
I'm unaware of what but maybe hot sauce?
Explanation:
Is the something missing from the question
